Chapter 5.3, Problem 39E

Let’s eat: A fast-food restaurant chain has 600 outlets in the United States. The following table categorizes them by city population size and location; and presents the number of restaurants in each category. A restaurant is to be chosen at random from the 600 to test market a new menu.

Given that the restaurant is located in a city with a population over 500,000, what is die probability that it is in the Northeast?

Given that the restaurant is located in the Southeast, what is the probability that it is in a city with a population under 50,000?

Given that the restaurant is located in the Southwest what is the probability that it is in a city with a population of 500,000 or less?

Given that the restaurant is located in a city with a population of 500,000 or less, what is the probability that it is in the Southwest?

Given that the restaurant is located in the South (either SE or SW), what is the probability that it is in a city with a population of 50,000 or more?

To determine

The probability that the restaurant is in Northeast, it is given that restaurant is located in a city with a population over 500,000.

Answer to Problem 39E

  $\frac{30}{53}$

Explanation of Solution

Given:

The following table categorizes 600 outlets of China restaurant by city population, size, and location.

Population Of City	NE	SE	SW	NW
Under 50,000	30	35	15	5
50,000-500,000	60	90	70	30
Over 500,000	150	25	30	60

Formula Used:

Probability of an event is given by

  $\frac{\text{Number}\text{of}\text{Favorable}\text{Outcome}}{\text{Number}\text{of}\text{total}\text{outcome}}$

For any two dependent events $A$ and B ,

  $P\left(B/A\right)=\frac{P\left(AandB\right)}{P\left(A\right)}$

Where

  $P\left(B/A\right)=$ The probability of occurring $B$ under the assumption that $A$ has been occurred.

Calculation:

Let $A$ be the event that restaurant is located in a city with a population over 500,000.

  $\begin{array}{c}n\left(A\right)=150+25+30+60\\ =265\\ n\left(S\right)=600\\ P\left(A\right)=\frac{265}{600}\end{array}$

Let $B$ be the event that the restaurant is in Northeast.

So, $A\text{and}B$ be the event that the restaurant is in Northeast and size of population over 500,000.

  $\begin{array}{c}n\left(AandB\right)=150\\ P\left(AandB\right)=\frac{150}{600}\end{array}$

Therefore, $B/A$ be the event that the restaurant is in Northeast, it is given that restaurant is located in a city with a population over 500,000.

So, the probability of the event “the restaurant is in Northeast, it is given that restaurant is located in a city with a population over 500,000” is calculated as follows:

  $\begin{array}{c}P\left(B/A\right)=\frac{P\left(AandB\right)}{P\left(A\right)}\\ =\frac{\frac{150}{600}}{\frac{265}{600}}\\ =\frac{150}{265}\\ =\frac{30}{53}\end{array}$

Hence, the probability of the event thatthe restaurant is in Northeast, it is given that restaurant is located in a city with a population over 500,000 is $\frac{30}{53}$ .

To determine

The probability restaurant is located in a city with a population under 50,000 it is given that the restaurant is located in the Southeast.

Answer to Problem 39E

  $\frac{7}{30}$

Explanation of Solution

Given:

The following table categorizes 600 outlets of China restaurant by city population, size, and location.

Population Of City	NE	SE	SW	NW
Under 50,000	30	35	15	5
50,000-500,000	60	90	70	30
Over 500,000	150	25	30	60

Formula Used:

Probability of an event is given by

  $\frac{\text{Number}\text{of}\text{Favorable}\text{Outcome}}{\text{Number}\text{of}\text{total}\text{outcome}}$

For any two dependent events $A$ and B ,

  $P\left(B/A\right)=\frac{P\left(AandB\right)}{P\left(A\right)}$

Where

  $P\left(B/A\right)=$ The probability of occurring $B$ under the assumption that $A$ has been occurred.

Calculation:

Let $C$ be the event that the restaurant is located in the Southeast.

  $\begin{array}{c}n\left(C\right)=35+25+90\\ =150\\ n\left(S\right)=600\\ P\left(C\right)=\frac{150}{600}\end{array}$

Let $D$ be the event that restaurant is located in a city with a population under 50,000.

So, $C\text{and}D$ be the event that the restaurant is in Southeast and size of population is under 50,000.

  $\begin{array}{c}n\left(CandD\right)=35\\ P\left(CandD\right)=\frac{35}{600}\end{array}$

Therefore, $D/C$ be the event that restaurant is located in a city with a population under 50,000 it is given that the restaurant is located in the Southeast.

So, the probability of the event “restaurant is located in a city with a population under 50,000 it is given that the restaurant is located in the Southeast” is calculated as follows:

  $\begin{array}{c}P\left(D/C\right)=\frac{P\left(CandD\right)}{P\left(C\right)}\\ =\frac{\frac{35}{600}}{\frac{150}{600}}\\ =\frac{35}{150}\\ =\frac{7}{30}\end{array}$

Hence, the probability restaurant is located in a city with a population under 50,000 it is given that the restaurant is located in the Southeastis $\frac{7}{30}$ .

To determine

The probability that the city has a population 500,000 or less, given that restaurant is located in Southwest.

Answer to Problem 39E

The probability that the city has a population 500,000 or less, given that restaurant is located in Southwest is $\frac{17}{23}$ .

Explanation of Solution

Given:

The following table categorizes 600 outlets of China restaurant by city population, size, and location.

Population Of City	NE	SE	SW	NW
Under 50,000	30	35	15	5
50,000-500,000	60	90	70	30
Over 500,000	150	25	30	60

Formula Used:

Probability of an event is given by

  $\frac{\text{Number}\text{of}\text{Favorable}\text{Outcome}}{\text{Number}\text{of}\text{total}\text{outcome}}$

For any two dependent events $A$ and B ,

  $P\left(B/A\right)=\frac{P\left(AandB\right)}{P\left(A\right)}$

Where

  $P\left(B/A\right)=$ The probability of occurring $B$ under the assumption that $A$ has been occurred.

Calculation:

Let $W$ be the event that the restaurant is located in the Southwest.

  $\begin{array}{c}n\left(W\right)=15+70+30\\ =115\\ n\left(S\right)=600\\ P\left(W\right)=\frac{115}{600}\end{array}$

Let $E$ be the event that restaurant is located in a city that has a population 500,000 or less.

So, $W\text{and}E$ be the event that the restaurant is in Southwast and size of population is 500,000 or less.

  $\begin{array}{c}n\left(WandE\right)=70+15\\ =85\\ P\left(WandE\right)=\frac{85}{600}\end{array}$

Therefore, $E/W$ be the event that the city has a population 500,000 or less, given that restaurant is located in Southwest.

So, the probability of the event that the city has a population 500,000 or less, given that restaurant is located in Southwest” is calculated as follows:

  $\begin{array}{c}P\left(E/W\right)=\frac{P\left(WandE\right)}{P\left(W\right)}\\ =\frac{\frac{85}{600}}{\frac{115}{600}}\\ =\frac{85}{115}\\ =\frac{17}{23}\end{array}$

Hence, the probability that the city has a population 500,000 or less, given that restaurant is located in Southwest is $\frac{17}{23}$ .

To determine

The probability that the restaurant is in Southwest, it is given that restaurant is located in a city with a population 500,000 or less.

Answer to Problem 39E

  $\frac{17}{67}$

Explanation of Solution

Given:

The following table categorizes 600 outlets of China restaurant by city population, size, and location.

Population Of City	NE	SE	SW	NW
Under 50,000	30	35	15	5
50,000-500,000	60	90	70	30
Over 500,000	150	25	30	60

Formula Used:

Probability of an event is given by

  $\frac{\text{Number}\text{of}\text{Favorable}\text{Outcome}}{\text{Number}\text{of}\text{total}\text{outcome}}$

For any two dependent events $A$ and B ,

  $P\left(B/A\right)=\frac{P\left(AandB\right)}{P\left(A\right)}$

Where

  $P\left(B/A\right)=$ The probability of occurring $B$ under the assumption that $A$ has been occurred.

Calculation:

Let $F$ be the event that restaurant is located in a city with a population 500,000 or less.

  $\begin{array}{c}n\left(F\right)=600-\left(150+25+30+60\right)\\ =600-265\\ =335\\ n\left(S\right)=600\\ P\left(F\right)=\frac{335}{600}\end{array}$

Let $S$ be the event that the restaurant is inSouthwest.

So, $S\text{and}F$ be the event that the restaurant is in Southwast and size of population is 500,000 or less.

  $\begin{array}{c}n\left(SandF\right)=70+15\\ =85\\ P\left(SandF\right)=\frac{85}{600}\end{array}$

Therefore, $S/F$ be the event that the restaurant is in Southwest, it is given that restaurant is located in a city with a population 500,000 or less.

So, the probability of the event “the restaurant is in Southwest, it is given that restaurant is located in a city with a population 500,000 or less” is calculated as follows:

  $\begin{array}{c}P\left(S/F\right)=\frac{P\left(SandF\right)}{P\left(F\right)}\\ =\frac{\frac{85}{600}}{\frac{335}{600}}\\ =\frac{85}{335}\\ =\frac{17}{67}\end{array}$

Hence, the probability of the event thatthe restaurant is in Southwest, it is given that restaurant is located in a city with a population 500,000 or less is $\frac{17}{67}$ .

To determine

The probability that the city has a population 50,000 or more, given that restaurant is located in South.

Answer to Problem 39E

The probability that the city has a population 50,000 or more, given that restaurant is located in Southis $\frac{43}{53}$ .

Explanation of Solution

Given:

The following table categorizes 600 outlets of China restaurant by city population, size, and location.

Population Of City	NE	SE	SW	NW
Under 50,000	30	35	15	5
50,000-500,000	60	90	70	30
Over 500,000	150	25	30	60

Formula Used:

Probability of an event is given by

  $\frac{\text{Number}\text{of}\text{Favorable}\text{Outcome}}{\text{Number}\text{of}\text{total}\text{outcome}}$

For any two dependent events $A$ and B ,

  $P\left(B/A\right)=\frac{P\left(AandB\right)}{P\left(A\right)}$

Where

  $P\left(B/A\right)=$ The probability of occurring $B$ under the assumption that $A$ has been occurred.

Calculation:

Let $X$ be the event that the restaurant is located in the South.

  $\begin{array}{c}n\left(X\right)=35+90+25+15+70+30\\ =265\\ n\left(S\right)=600\\ P\left(X\right)=\frac{265}{600}\end{array}$

Let $G$ be the event that restaurant is located in a city that has a population 50,000 or more.

So, $X\text{and}G$ be the event that the restaurant is in South and size of population is 50,000 or more.

  $\begin{array}{c}n\left(X\text{and}G\right)=90+70+25+30\\ =215\\ P\left(X\text{and}G\right)=\frac{215}{600}\end{array}$

Therefore, $G/X$ be the event that the city has a population 50,000 or more, given that restaurant is located in South.

So, the probability of the event that that “the city has a population 50,000 or more, given that restaurant is located in South” is calculated as follows:

  $\begin{array}{c}P\left(G/X\right)=\frac{P\left(GandX\right)}{P\left(X\right)}\\ =\frac{\frac{215}{600}}{\frac{265}{600}}\\ =\frac{215}{265}\\ =\frac{43}{53}\end{array}$

Hence, the probability that that the city has a population 50,000 or more, given that restaurant is located in South is $\frac{43}{53}$ .